Section 7.1 Introduction to Rational Expressions Copyright © 2013, 2009, and 2005 Pearson...

Section 7.1

Introduction to Rational

Expressions

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Basic Concepts

• Simplifying Rational Expressions

• Applications

Basic Concepts

Rational expressions can be written as quotients (fractions) of two polynomials.Examples include:

2

3

2

, , 5

4

6

3

4 1

4 8

x x

x

x

xx

Example

If possible, evaluate each expression for the given value of the variable.a. b. c.

Solutiona. b. c.

1; 3

3x

x

2

; 43 4

ww

w

4; 5

4

ww

w

1; 3

3x

x

2

;3 4

4w

ww

4;

45

ww

w

1

3

1

3 6

2( )

3

4

( 44)

16

12 4

16

28

4 ( )

5)

5

( 4

91

9

Example

Find all values of the variable for which each expression is undefined.a. b. c.

Solutiona. b. c.

2

1

x

2

4

w

w 2

6

4w

2

1

x

2

4

w

w 2

6

4w

Undefined when x2 = 0 or when x = 0.

Undefined when w – 4 = 0 or when w = 4.

Undefined when w2 – 4 = 0 or when w = 2.

Example

Simplify each fraction by applying the basic principle of fractions.a. b. c.

Solutiona. The GCF of 9 and 15 is 3.

b. The GCF of 20 and 28 is 4.

c. The GCF of 45 and 135 is 45.

9

15

20

28

45

135

9

15 5

3

3

3

3

5

20

28 7

4

4

5

5

7

45

135

145

5 34

1

3

Example

Simplify each expression.a. b. c.

Solutiona. b. c.

2

16

4

y

y

3 12

4 16

x

x

2

2

25

2 7 15

x

x x

2

16

4

y

y

3 12

4 16

x

x

44

4

y

y y

4

y

3( )

4(

4

)4

x

x

3

4

2

2

25

2 7 15

x

x x

( )( 5)

(2

5

3)( )5

x

x

x

x

5

2 3

x

x

Example

Simplify each expression.a. b.

Solutiona. b.

7

2 14

y

y

8

8

x

x

7

2 14

y

y

1( )

2(

7

7)

y

y

1

2

8

8

x

x

(8 )

8

x

x

8

8

x

x

81

8

x

x

Example

Suppose that n balls, numbered 1 to n, are placed in a container and two balls have the winning number.a. What is the probability of drawing the winning ball at random?b. Calculate this probability for n = 100, 1000 and 10,000.c. What happens to the probability of drawing the winning ball as the number of balls increases?Solutiona. There are 2 chances of drawing the winning ball. 2

n

Example (cont)

b. Calculate this probability for n = 100, 1000 and 10,000.

c. What happens to the probability of drawing the winning ball as the number of balls increases?

The probability decreases.

2 1

100 50

2 1

1000 500

2 1

10,000 5000

Section 7.2

Multiplication and Division of Rational

Expressions


Objectives

• Review of Multiplication and Division of Fractions

• Multiplication of Rational Expressions

• Division of Rational Expressions

Example

Multiply and simplify your answers to lowest terms. a. b. c.

Solutiona.

b.

c.

4 5

9 7

415

5 2 5

7 8

4 5 20

9 7 63

4 15 4 6015 12

5 1 5 5

2 5 5 2 5 1 5

7 8 7 8 7 4 28

Example

Divide and simplify your answers to lowest terms. a. b. c.

Solutiona. b.

c.

1 3

6 5

618

7 4 11

5 15

1 3

6 5

618

7

4 11

5 15

1 5

6 3

5

18

1 6

7 18

6 1

7 18

1

21

60

55

4 15

5 11

12

11

12 5

11 5

Example

Multiply and simplify to lowest terms. Leave your answers in factored form. a. b.

Solutiona. b.

2

6 5

10 12

x

x

3 4

2 1 3 9

x x

x x

2

6 5

10 12

x

x

210 12

6 5x

x

21

0

0

3

2

x

x

1

4x

3 4

2 1 3 9

x x

x x

( 3)( 4)

(2 1) 3 )9(x x

x x

( )( 4)

3(2 1)( 3)

3 x

x

x

x

4

3(2 1)

x

x

Example

Multiply and simplify to lowest terms. Leave your answer in factored form.

Solution

2

2

16 3

9 4

x x

x x

2

2

( 3)

(

( )

4)

6

( 9)

1

x

x

x

x

2

2

16 3

9 4

x x

x x

( 3)( 4)( 4)

( 3)( 3 4))(x

xx x

x x

( 4)( )( )

( 3)

3

( 3 )

4

4)(

x

x

x x

x x

( 4)

( 3)

x

x

Example

Divide and simplify to lowest terms.a. b.

Solutiona. b.

3 2 1

6

x

x x

2

2

16 4

2 8 2

x x

x x x

3 2 1

6

x

x x

3 6

2 1

x

x x

18

(2 1)

x

x x

18

2 1x

2

2

16

2 8

4

2

x

x x

x

x

2

2

16

42 8

2x

x x

x

x

( 4)( 4) 2

( 2)( 4) 4

x x x

x x x

( )( )( )

( )( )(

4

2 4

24

)4

x

xx

x

x

x

1

Section 7.3

Addition and Subtraction with

Like Denominators


Objectives

• Review of Addition and Subtraction of Fractions

• Rational Expressions Having Like Denominators

Example

Simplify each expression to lowest terms. a. b.

Solution

a.

b.

4 1

7 7

1 5

9 9

4 1

7 7 4 1

7

5

7

1 5

9 9 1 5

9

6

9

2

3

Example

Simplify each expression to lowest terms. a. b.

Solution

a.

b.

13 7

18 18

15 11

30 30

13 7

18 18

13 7

18

6

18

1

3

15 11

30 30

15 11

30

2

15

4

30

To add two rational expressions having like denominators, add their numerators. Keep the same denominator.

C is nonzero

SUMS OF RATIONAL EXPRESSIONS

A B A B

C C C

Example

Add and simplify to lowest terms.a. b.

Solutiona.

b.

4 1 2

3 3

x x

x x

2 2

5

7 10 7 10

x

x x x x

3

4

3

1 2

x x

x x

4 1 2

3

x x

x

3

5 1x

x

2 2

5

7 10 7 10

x

x x x x

2

5

7 10

x

x x

5

5 2

x

x x

1

2x

Example

Add and simplify to lowest terms.a. b.

Solutiona.

b.

7 4

ab ab

2 2 2 2

w y

w y w y

7 4

ab ab

7 4

ab

11

ab

2 2 2 2

w y

w y w y

2 2

w y

w y

( )( )

w y

w y w y

1

w y

To subtract two rational expressions having like denominators, subtract their numerators. Keep the same denominator.

C is nonzero

DIFFERENCES OF RATIONAL EXPRESSIONS

A B A B

C C C

Example

Subtract and simplify to lowest terms.a. b.

Solutiona.

b.

2 2

6 6x

x x

2 2

2 3 4

1 1

x x

x x

2 2

6 6x

x x

2

6 6x

x

2

6 6x

x

2

x

x

1

x

2 2

2 3 4

1 1

x x

x x

2

2 3 4

1

x x

x

1

1 1

x

x x

2

1

1

x

x

1

1x

Example

Subtract and simplify to lowest terms.

Solution

7 2

2 2

a a

a a

7 2

2 2

a a

a a

7 ( 2)

2

a a

a

6 2

2

a

a

Section 7.4

Addition and Subtraction with

Unlike Denominators


Objectives

• Finding Least Common Multiples

• Review of Fractions Having Unlike Denominators

• Rational Expressions Having Unlike Denominators

The least common multiple (LCM) of two or more polynomials can be found as follows.

Step 1: Factor each polynomial completely.

Step 2: List each factor the greatest number of times that it occurs in either

factorization.

Step 3: Find the product of this list of factors. The result is the LCM.

FINDING THE LEAST COMMON MULTIPLE

Example

Find the least common multiple of each pair of expressions.a. 6x, 9x4 b. x2 + 7x + 12, x2 + 8x + 16 SolutionStep 1: Factor each polynomial completely.

6x = 3 ∙ 2 ∙ x 9x4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x

Step 2: List each factor the greatest number of times.3 ∙ 3 ∙ 2 ∙ x ∙ x ∙ x ∙ x

Step 3: The LCM is 18x4.

Example (cont)

b. x2 + 7x + 12, x2 + 8x + 16

Step 1: Factor each polynomial completely.x2 + 7x + 12 = (x + 3)(x + 4)x2 + 8x + 16 = (x+ 4)(x + 4)

Step 2: List each factor the greatest number of times.(x + 3), (x + 4), and (x + 4)

Step 3: The LCM is (x + 3)(x + 4)2.

Example

Simplify each expression.a. b.

Solutiona. The LCD is the LCM, 42.

b. The LCD is 60.

4 1

7 6

5 11

12 30

4 1

7 6

4 6 1 7

7 6 6 7

24 7

42 42

31

42

5 11

12 30

5 5 11 2

12 5 30 2

25 22

60 60

3

60

1

20

Example

Find each sum and leave your answer in factored form.a. b.

Solutiona. The LCD is x2.

b.

2

2 5

x x

4 3

1 1x x

2

2 5

x x

2

2 5x

x x x

2 2

2 5x

x x

2

2 5x

x

4 3

1 1x x

4 3 1

1 1 1x x

4 3

1 1x x

1

1x

Example

Simply the expression. Write your answer in lowest terms and leave it in factored form. SolutionThe LCD is x(x + 7).

3 5

7

x

x x

3 5

7

x

x x

3 7 5

7 7

x x x

x x x x

3 7 5

7 7

x x x

x x x x

3 7 5

7

x x x

x x

2 4 21 5

7

x x x

x x

2 21

7

x x

x x

Example

Simplify the expression. Write your answer in lowest terms and leave it in factored form.Solution

2 2

6 5

6 9 9x x x

6 5

3 3 3 3x x x x

2 2

6 5

6 9 9x x x

6 5

3 3 3 33

33

3

x

x x xx x

x

x

6 3 5 3

3 3 3 3 3 3

x x

x x x x x x

6 18 5 15

3 3 3

x x

x x x

33

3 3 3

x

x x x

Example

Add and then find the reciprocal of the result.

SolutionThe LCD is RS.

1 1,

R S

1 1 1 1

R S R

R

S

S

S R

S R

RS RS

S R

RS

The reciprocal is

.RS

S R

Section 7.5

Complex Fractions


Objectives

• Basic Concepts

•Simplifying Complex Fractions

Basic Concepts

A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both.

Complex fractions involve division of fractions. When dividing two fractions, we can multiply the numerator by the reciprocal of the denominator.

, 0 and 1,xf x a a a For any real numbers a, b, c, and d,

where b, c, and d are nonzero.

SIMPLIFYING BASIC COMPLEX FRACTIONS

,

aa db

c b cd

Simplifying Complex Fractions (Method 1)

To simplify a complex fraction, perform the following steps.STEP 1: Write the numerator as a single fraction; write the denominator as a single fraction.

STEP 2: Divide the denominator into the numerator by multiplying the numerator and the reciprocal of the denominator.

STEP 3: Simplify the result to lowest terms.

Example

Simplify each complex fraction.a. b.

Solutiona.

b.

56

23

2

10

15

xy

xy

56

23

35

6 2

5

4

2

10

15

xy

xy

215

10

x y

y x

215

10

xy

xy 3

2

y

Example

Simplify. Write your answer in lowest terms.

Solution

13

13

3

3n

n

n

n

13

13

3

3n

n

n

n

2

2

9 13 3

9 13 3

nn n

nn n

2

2

9 13

9 13

nn

nn

2

2

9 1 3

3 9 1

n n

n n

2

2

9 1

9 1

n

n

Example

Simplify. Write your answer in lowest terms.1 1

2 2

2 24 4

x x

x x

For the numerator, the LCD is (x + 2)(x – 2).

For the denominator, the LCD is (x + 4)(x – 4).

2 22 2 2 2

2 4 2 4

4 4 4 4

x xx x x x

x x

x x x x

42 2

164 4

x x

x x

4 44

2 2 16

x x

x x

4 4

4 2 2

x x

x x

Simplifying Complex Fractions (Method II)

To simplify a complex fraction, perform the following steps.STEP 1: Find the LCD of all fractions within the complex fraction.

STEP 2: Multiply the numerator and the denominator of the complex fraction by the LCD.

STEP 3: Simplify the result to lowest terms.

Example

Simplify.

Solution

1 1x y

x y

1 1x y

x y

1 1x y xy

x y xy

xy xyx y

xy x y

y x

xy x y

1

xy

Example

Simplify.

Solution

641

811

x x

x x

641

811

x x

x x

641

811

1

1x x

x x

x x

x x

4 1 6 111 8 1

1

x x x xx xx x x xx x

4 6 1

8 1

x x

x x

4 6 6

8 8

x x

x x

2 6

9 8

x

x

Example

Simplify.

Solution

3 3

3 3

1 1

11x y

x y

3 3

3 3

1 1

11x y

x y

3 3

3 3

3 31 1

3 311

x y

x y

x y

x y

3 3

3 3 1

y x

x y

Section 7.6

Rational Equations and

Formulas


Objectives

• Solving Rational Equations

• Rational Expressions and Equations

• Graphical and Numerical Solutions

• Solving a Formula for a Variable

• Applications

Rational Equations

If an equation contains one or more rational expressions, it is called a rational equation.

Example

Solve each equation.a. b.

Solutiona. b.

9 4

7 x

6

2 1x

x

9 4

7 x

9 28x

28

9x

6

2 1x

x

6

2 1 1

x

x

2 1 6(1)x x

22 6x x 22 6 0x x

2 3 2 0x x

2 3 0x

3

2x 2x

The solutions are 3

2 and .2

The solution is 28

.9

2 0x

Example

Determine whether you are given an expression or an equation. If it is an expression, simplify it and then evaluate it for x = 4. If it is an equation, solve it.a. b.

Solutiona. There is an equal sign, so it is an equation.

24

3 3

x x

x x

2 3 10

2 2

x x

x x

23 4 3 3

3 3

x xx x x

x x

2 4 12x x x

4x 3 12x The answer checks.

The solution is −4.

Example (cont)

b. There is no equals sign, so it is an expression. The common denominator is x – 2, so we can add the numerators.

2 3 10

2

x x

x

2 3 10

2 2

x x

x x

2 5

2

x x

x

5x

When x = 4, the expression evaluates 4 + 5 = 9.

2 3 10

2 2

x x

x x

Example

Solve graphically and numerically.

SolutionGraph and

2

1x

x

1

2

1y

x

2 .y x

The solutions are −2 and 1.

x −3 −2 −1 0 1 2 3

−1 −2 −− 2 1 2

3

1

2(−2, −2)

(1, 1)

Example (cont)

Solve graphically and numerically.

SolutionNumerical Solution

2

1x

x

The solutions are −2 and 1.

x −3 −2 −1 0 1 2 3

−1 −2 −− 2 1

−3 −2 −1 0 1 2 3

1

2

1y

x

2

31

2

2y x

Example

Solve the equation for the specified variable.

Solution

2 for C r r

2 for C r r

2

2 2

rC

2

Cr

Example


Solution

2 for

A

h AB b

2h

b

A

B

( ) 2h B b A

( )

2

h B bA

Example


Solution

2 for S B sl s

2 for S B ls s 2 S B B B ls

2 S B ls 2

2 2

S B

l

sl

l

2

S Bs

l

Example

A pump can fill a swimming pool ¾ full in 6 hours, another can fill the pool ¾ full in 9 hours. How long would it take the pumps to fill the pool ¾ full, working together?Solution

16 9

t t

36 36(1)6 9

t t 6 4 36t t

6 33 3

10 5t

The two pumps can fill the pool ¾ full in hours.3

35

Section 7.7

Proportions and Variation


Objectives

• Proportions

• Direct Variation

• Inverse Variation

• Analyzing Data

• Joint Variation

Proportions

A proportion is a statement (equation) that two ratios (fractions) are equal.

The following property is a convenient way to solve proportions:

a

b d

c is equivalent to ,ad bc

provided b ≠ 0 and d ≠ 0.

Example

On an elliptical machine, Francis can burn 370 calories in 25 minutes. If he increases his work time to 30 minutes, how many calories will he burn?SolutionLet x be the equivalent amount of calories.

25 30

370 x

Minutes Minutes=

Calories Calories

25 11,100x

444x Thus, in 30 minutes, Francis will burn 444 calories.

Example

A 6-foot tall person casts a shadow that is 8-foot long. If a nearby tree casts a 32-foot long shadow, estimate the height of the tree.SolutionThe triangles are similar because the measures of its corresponding angles are equal. Therefore corresponding sides are proportional.

6 ft h

8 ft 32 ft

32

6 8

h

Height Shadow length=

Height Shadow length8 192h

24h The tree is 24 feet tall.

Example

Let y be directly proportional to x, or vary directly with x. Suppose y = 9 when x = 6. Find y when x = 13.SolutionStep 1 The general equation is y = kx. Step 2 Substitute 9 for y and 6 for x in y = kx. Solve for k.Step 3 Replace k with 9/6 in the equation y = 9x/6.Step 4 To find y, let x = 13.

9 6

9

6

y kx

k

k

9

69

(13)619.5

y x

y

y

Example

The table lists the amount of pay for various hours worked.

a. Find the constant of proportionality. b. Predict the pay for 19 hours of work.

Hours Pay

6 $138

11 $253

15 $345

23 $529

31 $713

6, 138

11, 253

15, 345

23, 529

31, 713

0

100

200

300

400

500

600

700

800

0 10 20 30 40P

ay

(do

llars

)

Hours

Example (cont)

The slope of the line equals the proportionality, k. If we use the first and last data points (6, 138) and (31, 713), the slope is

The amount of pay per hour is $23. The graph of the line y = 23x, models the given graph.To find the pay for 19 hours, substitute 19 for x.

713 138

31 6k

23

y = 23x,y = 23(19)y = 437

19 hours of work would pay $437.00

Example

Let y be inversely proportional to x, or vary inversely with x. Suppose y = 6 when x = 4. Find y when x = 8.SolutionStep 1 The general equation is y = k/x. Step 2 Substitute 6 for y and 4 for x in Solve for k.Step 3 Replace k with 24 in the equation y = k/x.Step 4 To find y, let x = 8.

64

24

ky

xk

k

24

83

ky

x

y

y

Example

Determine whether the data in each table represent direct variation, inverse variation, or neither. For direct and inverse variation, find the equation.a.

b.

c.

x 3 7 9 12

y 12 28 32 48

x 5 10 12 15

y 12 6 5 4

x 8 11 14 21

y 48 66 84 126

Neither the product xy nor the ratio y/x are constant in the data in the table. Therefore there is neither direct variation nor indirect variation in this table.

As x increases, y decreases. Because xy = 60 for each data point, the equation y = 60/x models the data. This represents an inverse variation.

The equation y = 6x models the data. The data represents direct variation.

, 0 and 1,xf x a a a

Let x, y, and z denote three quantities. Then z varies jointly with x and y if there is a nonzero number k such that

JOINT VARIATION

.z kxy

Example

The strength S of a rectangular beam varies jointly as its width w and the square of its thickness t. If a beam 5 inches wide and 2 inches thick supports 280 pounds, how much can a similar beam 4 inches wide and 3 inches thick support?SolutionThe strength of the beam is modeled by S = kwt2.

2280 5 2k

280

5 4k

14k

Example (cont)

Thus S = 14wt2 models the strength of this type of beam. When w = 4 and t = 3, the beam can support

S = 14 ∙ 4 ∙ 32 = 504 pounds

Section 7.1 Introduction to Rational Expressions Copyright © 2013, 2009, and 2005 Pearson...

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Transcript of Section 7.1 Introduction to Rational Expressions Copyright © 2013, 2009, and 2005 Pearson...